Problem: $\begin{cases}c(1)=6\\\\ c(n)=c(n-1)-16 \end{cases}$ Find the $3^{\text{rd}}$ term in the sequence.
This is a recursive formula. It tells us that the first term is $6$ and that the common difference is $-16$. $\begin{aligned} {c(1)}&=6 \\\\ {c(2)}&={c(1)}-16=-10 \\\\ {c(3)}&={c(2)}-16=-26 \end{aligned}$ The $3^{\text{rd}}$ term is $-26$.